What is the derivative of #sin(sin(cos(x)sin(x)))#?

1 Answer
Sep 27, 2016

Answer:

#=cos(sin(cos(x)sin(x)))(cos(cos(x)sin(x)))cos (2x)#

Explanation:

I assume that x is in radian measure so that

#(sin x)'=cos x and (cos x)'=-sin x#.

Applying chain rule,

#(sin(sin(cos(x)sin(x)))'#

#=cos(sin(cos(x)sin(x)))(sin((cos(x)sin(x))'#

#=cos(sin(cos(x)sin(x)))(cos(cos(x)sin(x)))(cos(x)(sin(x))'+(cos(x))'sin(x))#

#=cos(sin(cos(x)sin(x)))(cos(cos(x)sin(x)))(cos^2(x)-sin^2(x))#

#=cos(sin(cos(x)sin(x)))(cos(cos(x)sin(x)))cos (2x)#